a+b+c \right )\left (
a^{2}+ b^{2} + c^{2} -ab-bc-ca \right )$ $2/\frac{
a^{3}+ b^{3} + c^{3} -3abc}{
a+b+c }\geq 0$
Lời giải
$1/$Ta có:
$a^{3}+ b^{3}=
\left ( a+b \right )^{3}-3ab
\left ( a+b \right )$
$\Rightarrow
a^{3}+ b^{3} + c^{3} -3abc=
\left ( a+b \right )^{3} + c^{3}
-3ab \left ( a+b+c \right ) $
$=
\left ( a+b+c \right ) [
\left ( a+b \right )^{2}-
\left ( a+b \right )c+
c^{2}-3ab ] $
$=
\left ( a+b+c \right ) \left ( a^{2} +
b^{2} + c^{2} -ab-bc-ca\right )$
$\Rightarrow
a^{3}+ b^{3} + c^{3} =
3abc+ \left ( a+b +c\right )
\left ( a^{2} + b^{2} + c^{2} -ab-bc-ca\right ) $
$2/$Theo chứng minh $1/$, ta có:
$ \frac{
a^{3}+ b^{3} + c^{3} -3abc}{a+b+c}=
a^{2} + b^{2} + c^{2} -ab-bc-ca $
$=\frac{1}{2}[
\left ( a-b \right )^{2}+
\left ( b-c \right )^{2 }
+ \left ( c-a \right )^{2} ]\geq 0 $
(Chú ý: vì $
a^{2}+ b^{2} + c^{2} \geq ab+bc+ca$, nên: $\frac{
a^{3}+ b^{3} + c^{3} -3abc }{a+b+c}\geq 0$
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Chuyên mục: Bất đẳng thức cơ bản
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